Exercise 1.2.1

Proof.

(a)

If $p=0$, by (1.7) and Ex. 1.1.3, $z_1=0,z_2=\sqrt[3]{-q}$, so $z_1,z_2$ are solutions of the equation $$0=z^6+q{z}^3-\frac{p^3}{27}=z^6+q{z}^3=z^3(z^3+q),$$

and the solutions of the cubic resolvent are $z_1=\omega z_1={\omega }^2{z}_1=0,z_2,\omega z_2,{\omega }^2{z}_2$.

We suppose now that $p\ne 0$.

By definition $z_1,z_2$ are such that

$$\begin{array}{lll}\hfill & z_1{z}_2=-\frac{p}{3},& \hfill \text{(1)}\\ \hfill & z_1^6+q{z}_1^3-\frac{p^3}{27}=0.& \hfill \text{(2)}\end{array}$$

Let $Z_1=z_1^3,Z_2=z_2^3$. Then

$$\begin{array}{lll}\hfill & Z_1{Z}_2=-\frac{p^3}{27}& \hfill \text{(3)}\\ \hfill & Z_1^2+q{Z}_1-\frac{p^3}{27}=0& \hfill \text{(4)}\end{array}$$

As $p\ne 0$, then $Z_1\ne 0$ by (4), and using (3) and (4),

$$Z_1+Z_2=Z_1-\frac{p^3}{27Z_1}=\frac{1}{Z_1}\left(Z_1^2-\frac{p^3}{27}\right)=\frac{1}{Z_1}(-q{Z}_1)=-q.$$

Thus, for all $Z\in ℂ$,

$$\begin{array}{lll}\hfill (Z-Z_1)(Z-Z_2)=Z^2-(Z_1+Z_2)Z+Z_1{Z}_2=Z^2+\mathit{qZ}-\frac{p^3}{27},& & \hfill \text{(5)}\end{array}$$

and so

$$Z_2^2+q{Z}_2-\frac{p^3}{27}=0,$$

$$z_2^6+q{z}_2^3-\frac{p^3}{27}=0.$$

$z_2$ is a root of the resolvent equation.

By (5) the resolvent equation is

$$\begin{array}{lll}\hfill 0=z^6+q{z}^3-\frac{p^3}{27}=(z^3-z_1^3)(z^3-z_2^3).& & \hfill \text{(6)}\end{array}$$

The solutions of this equation are so $z_1,z_2,\omega z_1,\omega z_2,{\omega }^2{z}_1,{\omega }^2{z}_2$.

(b)

By 1.2 A : $$\begin{array}{llll}\hfill & z_1=\frac{1}{3}(x_1+{\omega }^2{x}_2+\omega x_3),& \hfill & \\ \hfill & z_2=\frac{1}{3}(x_1+{\omega }^2{x}_3+\omega x_2).& \hfill & \end{array}$$

Multiplying by $\omega$, and by ${\omega }^2$ :

$$\begin{array}{llll}\hfill & \omega z_1=\frac{1}{3}(x_2+{\omega }^2{x}_3+\omega x_1),& \hfill & \\ \hfill & \omega z_2=\frac{1}{3}(x_3+{\omega }^2{x}_2+\omega x_1),& \hfill & \\ \hfill & {\omega }^2{z}_1=\frac{1}{3}(x_3+{\omega }^2{x}_1+\omega x_2),& \hfill & \\ \hfill & {\omega }^2{z}_2=\frac{1}{3}(x_2+{\omega }^2{x}_1+\omega x_2).& \hfill & \end{array}$$